Answer
$F^{'}(x)=\sqrt {x^{4}+1}$
Work Step by Step
$F(x)=\int_{-1}^{x}\sqrt {t^{4}+1}dt$
$F^{'}(x)=\frac{d}{dx}\int_{-1}^{x}\sqrt {t^{4}+1}dt$
$=\sqrt {x^{4}+1}$
This is valid since the function, $f(t)=\sqrt {t^{4}+1}$ is a continuous function over the implied domain, $\mathbb{R}$
